Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z^2 - 6z - 16}{z - 8} \times \dfrac{7z - 14}{z + 2} $
Solution: First factor the quadratic. $r = \dfrac{(z + 2)(z - 8)}{z - 8} \times \dfrac{7z - 14}{z + 2} $ Then factor out any other terms. $r = \dfrac{(z + 2)(z - 8)}{z - 8} \times \dfrac{7(z - 2)}{z + 2} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (z + 2)(z - 8) \times 7(z - 2) } { (z - 8) \times (z + 2) } $ $r = \dfrac{ 7(z + 2)(z - 8)(z - 2)}{ (z - 8)(z + 2)} $ Notice that $(z - 8)$ and $(z + 2)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 7\cancel{(z + 2)}(z - 8)(z - 2)}{ (z - 8)\cancel{(z + 2)}} $ We are dividing by $z + 2$ , so $z + 2 \neq 0$ Therefore, $z \neq -2$ $r = \dfrac{ 7\cancel{(z + 2)}\cancel{(z - 8)}(z - 2)}{ \cancel{(z - 8)}\cancel{(z + 2)}} $ We are dividing by $z - 8$ , so $z - 8 \neq 0$ Therefore, $z \neq 8$ $r = 7(z - 2) ; \space z \neq -2 ; \space z \neq 8 $